But we can't say Kx = mv^2/(l+x). That's what is my doubt.
Btw your answer is also wrong. Here we need to conserve angular momentum and energy
After you find velocity at max elongation, just try to find the radius of curvature at that instant, it won't come (l+x)

After you find velocity at max elongation, just try to find the radius of curvature at that instant, it won't come (l+x)

How would you do this ?
It is evident that the velocity won't be perpendicular to the length of the spring
But how do we find the angular momentum at the final position then ?@pratyaksh_tyagi@Chirag_Hegde

Here the radius of curvature at r=x is not l+x because at maximum elongation, the rate of change of radial speed
is non zero. When this rate of change of radial speed is ignored, the total acceleration in the radial direction becomes only centripetal. In this case the radius of curvature will be l+x.